Hi everyone.
I can't show this exercise:
Let $\displaystyle a,b\in{\mathbb{Z}$ be and $\displaystyle a<b$ then $\displaystyle a+1\leq{b}$
I've tried by contraction, but I haven't gotten anything.
Thanks
Well, I've done following:
Suppose that $\displaystyle a+1>b$, but $\displaystyle a<b\Leftrightarrow{a+1<b+1}$ hence it's a contraction because $\displaystyle a+1>b$ and $\displaystyle a+1<b+1$.
Therefore $\displaystyle a+1\leq{b}$
Is it correct?
It is difficult to answer that. We don't even know what set of axioms are in use here.
Here is a standard theorem in the development of number systems.
Given any two numbers, if $\displaystyle b-a>1$ then $\displaystyle (\exists j\in\mathbb{Z})[a<j<b].$
That depends upon Archimedes' principle and well ordering.